At a certain instant a rigid wheel is spinning about its center of mass with angular velocity of magnitude ω and angular acceleration of magnitude α. Consider a point a distance r from the axis of rotation. What would be the magnitude of the acceleration of this point on the wheel?

a = r√(w⁴ + (alpha)²)

Explanation:

let w be the magnitude of angular velocity

r = radius

v = velocity = wr

angular acceleration, ā = v²/r = (wr)²/r = w²r

Also,

â = dv/dt = rdw/dt

given : alpha = dw/dt

hence, â = (alpha)r

the resultant acceleration, a, by Pythagoras is given as

a = √( â² + ā²) = √(r²w⁴ + r²(alpha)²)

simplifying,

a = √r²(w⁴ + (alpha)²)

a = r√(w⁴ + (alpha)²)

i don't know that i agree with that but what's the question ?

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.